Area bounded by region of single curve Questions in English

(A) The area $A$ of the region bounded by the curve $y^2=x$,the $X$-axis,and the lines $x=1$ and $x=4$ in the first quadrant is given by the definite integral:
$A = \int_{1}^{4} y \, dx$
Since $y^2=x$ and we are in the first quadrant,$y = \sqrt{x} = x^{1/2}$.
$A = \int_{1}^{4} x^{1/2} \, dx$
Using the power rule for integration $\int x^n \, dx = \frac{x^{n+1}}{n+1}$:
$A = \left[ \frac{x^{3/2}}{3/2} \right]_{1}^{4} = \left[ \frac{2}{3} x^{3/2} \right]_{1}^{4}$
$A = \frac{2}{3} (4^{3/2} - 1^{3/2})$
$A = \frac{2}{3} (8 - 1) = \frac{2}{3} (7) = \frac{14}{3}$ sq. units.
Thus,the correct option is $A$.

(B) The given curve is $y^2 = 4x$,which implies $x = \frac{y^2}{4}$.
Since the region is bounded by the $Y$-axis $(x = 0)$,the curve $x = \frac{y^2}{4}$,and the line $y = 3$,the area $A$ is given by the integral with respect to $y$ from $y = 0$ to $y = 3$.
$A = \int_{0}^{3} x \, dy = \int_{0}^{3} \frac{y^2}{4} \, dy$.
$A = \frac{1}{4} \left[ \frac{y^3}{3} \right]_{0}^{3}$.
$A = \frac{1}{4} \left( \frac{3^3}{3} - 0 \right) = \frac{1}{4} \times \frac{27}{3} = \frac{9}{4}$ sq. units.
Therefore,the correct option is $B$.

(A) The given equation of the parabola is $x^2 = 12y$. Comparing this with $x^2 = 4ay$,we get $4a = 12$,which implies $a = 3$.
The focus of the parabola is $(0, a) = (0, 3)$.
The equation of the latus rectum is $y = 3$.
The region is bounded by the parabola $y = \frac{x^2}{12}$ and the line $y = 3$.
The points of intersection are found by substituting $y = 3$ into $x^2 = 12y$,giving $x^2 = 36$,so $x = \pm 6$.
The area $A$ is given by the integral $A = \int_{-6}^{6} (3 - \frac{x^2}{12}) dx$.
Since the function is even,$A = 2 \int_{0}^{6} (3 - \frac{x^2}{12}) dx$.
$A = 2 [3x - \frac{x^3}{36}]_{0}^{6} = 2 [3(6) - \frac{216}{36}] = 2 [18 - 6] = 2(12) = 24$ sq. units.

(B) The area $A$ is given by the integral of the absolute value of the function $y = \sin 2x$ from $x = 0$ to $x = \pi$.
$A = \int_{0}^{\pi} |\sin 2x| \, dx$
Since $\sin 2x$ changes sign at $x = \frac{\pi}{2}$,we split the integral:
$A = \int_{0}^{\pi/2} \sin 2x \, dx + \int_{\pi/2}^{\pi} -\sin 2x \, dx$
Evaluating the first part:
$\int_{0}^{\pi/2} \sin 2x \, dx = [-\frac{\cos 2x}{2}]_{0}^{\pi/2} = -\frac{1}{2}(\cos \pi - \cos 0) = -\frac{1}{2}(-1 - 1) = 1$
Evaluating the second part:
$\int_{\pi/2}^{\pi} -\sin 2x \, dx = [\frac{\cos 2x}{2}]_{\pi/2}^{\pi} = \frac{1}{2}(\cos 2\pi - \cos \pi) = \frac{1}{2}(1 - (-1)) = 1$
Total Area $A = 1 + 1 = 2$ sq. units.

(B) The equation of the ellipse is $2x^2 + 3y^2 = 1$.
This can be rewritten in the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ as:
$\frac{x^2}{1/2} + \frac{y^2}{1/3} = 1$.
Here,$a^2 = \frac{1}{2}$ and $b^2 = \frac{1}{3}$,so $a = \frac{1}{\sqrt{2}}$ and $b = \frac{1}{\sqrt{3}}$.
The area of an ellipse is given by the formula $A = \pi ab$.
Substituting the values,we get $A = \pi \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{3}} = \frac{\pi}{\sqrt{6}}$ sq. units.
Therefore,the correct option is $B$.

(C) The given equation is $|x| + y = 1$,which can be written as $y = 1 - |x|$.
This equation represents two lines:
$1$) For $x \ge 0$,$y = 1 - x$.
$2$) For $x < 0$,$y = 1 - (-x) = 1 + x$.
The region is bounded by these lines and the $x$-axis $(y = 0)$.
The vertices of the region are $(0, 1)$,$(1, 0)$,and $(-1, 0)$.
This forms a triangle with base length $b = 1 - (-1) = 2$ and height $h = 1$.
The area of the triangle is given by $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 1 = 1$ sq. units.

(C) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
For $a > b$,the eccentricity $e$ is given by $b^2 = a^2(1 - e^2)$,so $b^2 = a^2 - a^2e^2$,which implies $ae = \sqrt{a^2 - b^2}$.
The latus rectum is the line $x = ae$.
The area of the region bounded by the ellipse and its latus rectum is given by $2 \int_{ae}^{a} y \, dx$.
Since $y = \frac{b}{a} \sqrt{a^2 - x^2}$,the area is $2 \frac{b}{a} \int_{ae}^{a} \sqrt{a^2 - x^2} \, dx$.
Using the integral formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a})$,we evaluate from $ae$ to $a$:
Area $= \frac{2b}{a} [\frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a})]_{ae}^{a}$.
At $x = a$: $\frac{a}{2}(0) + \frac{a^2}{2} \sin^{-1}(1) = \frac{a^2}{2} \cdot \frac{\pi}{2} = \frac{a^2 \pi}{4}$.
At $x = ae$: $\frac{ae}{2} \sqrt{a^2 - a^2e^2} + \frac{a^2}{2} \sin^{-1}(e) = \frac{ae}{2} \cdot b + \frac{a^2}{2} \sin^{-1}(e)$.
Area $= \frac{2b}{a} [(\frac{a^2 \pi}{4}) - (\frac{aeb}{2} + \frac{a^2}{2} \sin^{-1}(e))]$.
This simplifies to the area bounded by the latus rectum and the vertex. However,standard problems often refer to the area between the latus rectum and the center or the total area. Given the options,the correct expression is $b(be + a \sin^{-1} e)$.

(A) The area $A$ of the region bounded by the curve $y = f(x)$,the $x$-axis,and the lines $x = a$ and $x = b$ is given by the formula:
$A = \int_{a}^{b} |f(x)| \, dx$
Given the curve $xy = 4$,we have $y = \frac{4}{x}$.
The region is bounded by $x = 1$ and $x = 3$.
Therefore,the area $A$ is:
$A = \int_{1}^{3} \frac{4}{x} \, dx$
$A = 4 \int_{1}^{3} \frac{1}{x} \, dx$
$A = 4 [\ln |x|]_{1}^{3}$
$A = 4 (\ln 3 - \ln 1)$
Since $\ln 1 = 0$,we get:
$A = 4 \ln 3$
Thus,the area is $4 \log 3$ sq. units.

(A) The equation of the parabola is $y^2 = 8x$. Comparing this with $y^2 = 4ax$,we get $4a = 8$,which implies $a = 2$.
The focus of the parabola is $(a, 0) = (2, 0)$.
The equation of the latus rectum is $x = a = 2$.
The parabola is symmetric about the $x$-axis.
The area of the region bounded by the parabola and its latus rectum is given by $A = 2 \int_{0}^{2} y \, dx$.
Since $y^2 = 8x$,we have $y = \sqrt{8x} = 2\sqrt{2} \sqrt{x}$.
Substituting this into the integral,we get $A = 2 \int_{0}^{2} 2\sqrt{2} \sqrt{x} \, dx = 4\sqrt{2} \int_{0}^{2} x^{1/2} \, dx$.
Evaluating the integral: $A = 4\sqrt{2} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{2} = 4\sqrt{2} \times \frac{2}{3} \times (2)^{3/2} = \frac{8\sqrt{2}}{3} \times 2\sqrt{2} = \frac{16 \times 2}{3} = \frac{32}{3}$ sq. units.

(C) The area $A$ is given by the integral of the absolute value of the function: $A = \int_{0}^{3\pi/2} |\cos x| \, dx$.
Since $\cos x$ is positive in $[0, \pi/2]$ and negative in $[\pi/2, 3\pi/2]$,we split the integral:
$A = \int_{0}^{\pi/2} \cos x \, dx + \int_{\pi/2}^{3\pi/2} -\cos x \, dx$.
Evaluating the first part: $[\sin x]_{0}^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1$.
Evaluating the second part: $-[\sin x]_{\pi/2}^{3\pi/2} = -(\sin(3\pi/2) - \sin(\pi/2)) = -(-1 - 1) = -(-2) = 2$.
Total area $A = 1 + 2 = 3$ square units.

(C) The curve is given by $y = x|x|$.
We can define this piecewise as:
$y = \begin{cases} x^2, & \text{if } x \ge 0 \\ -x^2, & \text{if } x < 0 \end{cases}$
The area $A$ is given by the integral $\int_{-1}^{1} |y| \, dx$.
Since the function $y = x|x|$ is an odd function,the area bounded by the curve and the $x$-axis from $x=-1$ to $x=1$ is calculated as:
$A = \int_{-1}^{0} | -x^2 | \, dx + \int_{0}^{1} | x^2 | \, dx$
$A = \int_{-1}^{0} x^2 \, dx + \int_{0}^{1} x^2 \, dx$
$A = [\frac{x^3}{3}]_{-1}^{0} + [\frac{x^3}{3}]_{0}^{1}$
$A = (0 - (-\frac{1}{3})) + (\frac{1}{3} - 0)$
$A = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$ square units.

(B) The curve is given by $y = x|x|$.
Since the interval is $x \in [0, 1]$,we have $x \ge 0$,so $|x| = x$.
Thus,the curve becomes $y = x \cdot x = x^2$ for $x \in [0, 1]$.
The area $A$ bounded by the curve,the $X$-axis,and the lines $x=0$ and $x=1$ is given by the definite integral:
$A = \int_{0}^{1} |y| \, dx = \int_{0}^{1} x^2 \, dx$.
Evaluating the integral:
$A = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} - 0 = \frac{1}{3}$.
Therefore,the area is $\frac{1}{3}$ sq. units.

(C) The given curve is $y^2 = 4x$,which implies $x = \frac{y^2}{4}$.
We need to find the area bounded by this curve,the $Y$-axis $(x = 0)$,and the line $y = 3$.
The region is bounded from $y = 0$ to $y = 3$.
The area $A$ is given by the integral $\int_{0}^{3} x \, dy$.
Substituting $x = \frac{y^2}{4}$,we get $A = \int_{0}^{3} \frac{y^2}{4} \, dy$.
$A = \frac{1}{4} \left[ \frac{y^3}{3} \right]_{0}^{3}$.
$A = \frac{1}{4} \left( \frac{27}{3} - 0 \right) = \frac{1}{4} \times 9 = \frac{9}{4}$ sq. units.
Therefore,the correct option is $C$.

(B) The area $A$ is given by the integral of the absolute value of the function over the given interval:
$A = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} |\cos x| \, dx$
In the interval $(\frac{\pi}{2}, \frac{3\pi}{2})$,$\cos x$ is negative in the interval $(\frac{\pi}{2}, \frac{3\pi}{2})$ except at the boundaries. Specifically,$\cos x \ge 0$ for $x \in [\frac{\pi}{2}, \frac{\pi}{2}]$ (not applicable) and $\cos x \le 0$ for $x \in [\frac{\pi}{2}, \frac{3\pi}{2}]$.
Thus,$|\cos x| = -\cos x$ for $x \in [\frac{\pi}{2}, \frac{3\pi}{2}]$.
$A = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} -\cos x \, dx$
$A = -[\sin x]_{\frac{\pi}{2}}^{\frac{3\pi}{2}}$
$A = -(\sin(\frac{3\pi}{2}) - \sin(\frac{\pi}{2}))$
$A = -(-1 - 1) = -(-2) = 2$
Therefore,the area is $2$ sq. units.

(C) The equation of the ellipse is given by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a^2 = 16$ and $b^2 = 9$.
This implies $a = 4$ and $b = 3$.
The formula for the area enclosed by an ellipse is $A = \pi ab$.
Substituting the values of $a$ and $b$,we get $A = \pi \times 4 \times 3 = 12 \pi$.
Therefore,the correct option is $C$.

(C) The given equation of the ellipse is $9x^2 + 4y^2 = 36$.
Dividing both sides by $36$, we get:
$\frac{9x^2}{36} + \frac{4y^2}{36} = 1$
$\frac{x^2}{4} + \frac{y^2}{9} = 1$
This is of the form $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$, where $a^2 = 9$ and $b^2 = 4$.
So, $a = 3$ and $b = 2$.
The area of an ellipse is given by the formula $A = \pi ab$.
Substituting the values, we get:
$A = \pi \times 3 \times 2 = 6\pi$ sq. units.

(B) The equation of the parabola is $y^2 = 12x$. Comparing this with $y^2 = 4ax$,we get $4a = 12$,so $a = 3$.
The latus rectum is the line $x = a = 3$.
The area bounded by the parabola and its latus rectum is given by $A = 2 \int_{0}^{a} y \, dx = 2 \int_{0}^{3} \sqrt{12x} \, dx$.
$A = 2 \times \sqrt{12} \int_{0}^{3} x^{1/2} \, dx = 2 \times 2\sqrt{3} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{3}$.
$A = 4\sqrt{3} \times \frac{2}{3} \times (3)^{3/2} = \frac{8\sqrt{3}}{3} \times 3\sqrt{3} = 8 \times 3 = 24$ sq. units.

(B) The given curve is $y^2 = 4x$,which implies $x = \frac{y^2}{4}$.
Since the region is bounded by the $Y$-axis $(x = 0)$,the curve $x = \frac{y^2}{4}$,and the line $y = 3$,the area $A$ is given by the integral with respect to $y$ from $y = 0$ to $y = 3$.
$A = \int_{0}^{3} x \, dy = \int_{0}^{3} \frac{y^2}{4} \, dy$
$A = \frac{1}{4} \left[ \frac{y^3}{3} \right]_{0}^{3}$
$A = \frac{1}{4} \left( \frac{3^3}{3} - 0 \right) = \frac{1}{4} \times \frac{27}{3} = \frac{9}{4}$ sq. units.
Therefore,the correct option is $B$.

(A) The area $A$ is given by the integral of the absolute value of the function $y = \cos x$ from $x = -\frac{\pi}{2}$ to $x = \pi$.
$A = \int_{-\frac{\pi}{2}}^{\pi} |\cos x| \, dx$.
Since $\cos x \ge 0$ for $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ and $\cos x \le 0$ for $x \in [\frac{\pi}{2}, \pi]$,we split the integral:
$A = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x \, dx + \int_{\frac{\pi}{2}}^{\pi} (-\cos x) \, dx$.
Evaluating the first part:
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x \, dx = [\sin x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2}) = 1 - (-1) = 2$.
Evaluating the second part:
$\int_{\frac{\pi}{2}}^{\pi} -\cos x \, dx = [-\sin x]_{\frac{\pi}{2}}^{\pi} = -(\sin \pi - \sin \frac{\pi}{2}) = -(0 - 1) = 1$.
Total area $A = 2 + 1 = 3$ sq. units.

(A) The equation of the parabola is $y^2 = 4ax$,where $4a = 4$,so $a = 1$.
The focus of the parabola is $(a, 0) = (1, 0)$.
The latus rectum is the line $x = 1$.
The area bounded by the parabola and the latus rectum is given by $A = 2 \int_{0}^{1} y \, dx$.
Since $y^2 = 4x$,we have $y = 2\sqrt{x}$.
$A = 2 \int_{0}^{1} 2\sqrt{x} \, dx = 4 \int_{0}^{1} x^{1/2} \, dx$.
$A = 4 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1} = 4 \times \frac{2}{3} [x^{3/2}]_{0}^{1}$.
$A = \frac{8}{3} [1 - 0] = \frac{8}{3}$ sq. units.

(C) The area $A$ is given by the integral of the absolute value of the function over the interval $[0, 2]$.
$A = \int_{0}^{2} |\sin(\pi x)| \, dx$
Since $\sin(\pi x) \ge 0$ for $x \in [0, 1]$ and $\sin(\pi x) \le 0$ for $x \in [1, 2]$,we split the integral:
$A = \int_{0}^{1} \sin(\pi x) \, dx + \int_{1}^{2} -\sin(\pi x) \, dx$
Evaluating the first part:
$\int_{0}^{1} \sin(\pi x) \, dx = [-\frac{\cos(\pi x)}{\pi}]_{0}^{1} = -\frac{1}{\pi}(\cos(\pi) - \cos(0)) = -\frac{1}{\pi}(-1 - 1) = \frac{2}{\pi}$
Evaluating the second part:
$\int_{1}^{2} -\sin(\pi x) \, dx = [\frac{\cos(\pi x)}{\pi}]_{1}^{2} = \frac{1}{\pi}(\cos(2\pi) - \cos(\pi)) = \frac{1}{\pi}(1 - (-1)) = \frac{2}{\pi}$
Total Area $A = \frac{2}{\pi} + \frac{2}{\pi} = \frac{4}{\pi}$ sq. units.
Therefore,the correct option is $C$.

(A) The area $A$ of the region bounded by the curve $y=f(x)$,the $X$-axis,and the lines $x=a$ and $x=b$ is given by the integral $A = \int_{a}^{b} |f(x)| dx$.
Here,the curve is $y = 9 - x^2$,and the limits are $x=0$ to $x=3$.
Since $9 - x^2 \geq 0$ for $x \in [0, 3]$,the area is given by:
$A = \int_{0}^{3} (9 - x^2) dx$
$A = [9x - \frac{x^3}{3}]_{0}^{3}$
$A = (9(3) - \frac{3^3}{3}) - (9(0) - \frac{0^3}{3})$
$A = (27 - \frac{27}{3}) - 0$
$A = 27 - 9 = 18$ sq. units.
Therefore,the correct option is $A$.

(A) The equation of the ellipse is $\frac{x^2}{16}+\frac{y^2}{9}=1$,which can be written as $\frac{x^2}{4^2}+\frac{y^2}{3^2}=1$. Here,$a=4$ and $b=3$.
From the figure,the shaded region is bounded by the ellipse in the first quadrant,specifically between $x=0$ and $x=4$,and the line connecting $(0,3)$ and $(4,0)$.
The equation of the line passing through $(0,3)$ and $(4,0)$ is $\frac{x}{4}+\frac{y}{3}=1$,which gives $y = \frac{3}{4}(4-x)$.
The area of the shaded region is the area under the ellipse minus the area under the line in the first quadrant.
Area $= \int_{0}^{4} \frac{3}{4}\sqrt{16-x^2} \, dx - \int_{0}^{4} \frac{3}{4}(4-x) \, dx$.
$= \frac{3}{4} \left[ \frac{x}{2}\sqrt{16-x^2} + \frac{16}{2}\sin^{-1}(\frac{x}{4}) \right]_{0}^{4} - \frac{3}{4} \left[ 4x - \frac{x^2}{2} \right]_{0}^{4}$.
$= \frac{3}{4} [0 + 8(\frac{\pi}{2})] - \frac{3}{4} [16 - 8] = 3\pi - 6 = 3(\pi-2)$ sq. units.

(C) The given equation of the curve is $x + 2y + 8 = 0$,which can be rewritten as $x = -2y - 8$.
The area $A$ of the region bounded by the curve $x = f(y)$ and the lines $y = c$ and $y = d$ is given by the integral $A = \int_{c}^{d} |x| \, dy$.
Here,$c = -3$ and $d = -1$.
So,$A = \int_{-3}^{-1} |-2y - 8| \, dy$.
Since for $y \in [-3, -1]$,the value of $-2y - 8$ is negative (e.g.,at $y = -2$,$-2(-2) - 8 = 4 - 8 = -4$),we have $|-2y - 8| = -(-2y - 8) = 2y + 8$.
Thus,$A = \int_{-3}^{-1} (2y + 8) \, dy$.
Integrating,we get $A = [y^2 + 8y]_{-3}^{-1}$.
Evaluating at the limits: $A = ((-1)^2 + 8(-1)) - ((-3)^2 + 8(-3))$.
$A = (1 - 8) - (9 - 24) = -7 - (-15) = -7 + 15 = 8$.
Therefore,the area is $8$ sq. units.

(B) The area $A$ of the region bounded by the parabola $y^2 = 4ax$ and its latus rectum $x = a$ is given by the integral:
$A = 2 \int_{0}^{a} \sqrt{4ax} \, dx$
$A = 2 \times 2\sqrt{a} \int_{0}^{a} x^{1/2} \, dx$
$A = 4\sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{a}$
$A = 4\sqrt{a} \times \frac{2}{3} \times a^{3/2} = \frac{8}{3} a^2$
Given that the area is $24$ sq. units,we have:
$\frac{8}{3} a^2 = 24$
$a^2 = 24 \times \frac{3}{8} = 9$
$a = \pm 3$
Thus,the correct option is $B$.

(D) The area $A$ is given by the integral of the absolute value of the function $y = x^2 - x - 6$ from $x = -1$ to $x = 1$.
First,we check if the curve crosses the $x$-axis in the interval $[-1, 1]$.
Setting $y = 0$,we get $x^2 - x - 6 = 0$,which factors as $(x - 3)(x + 2) = 0$.
The roots are $x = 3$ and $x = -2$.
Since neither $3$ nor $-2$ lies in the interval $[-1, 1]$,the function $y = x^2 - x - 6$ does not change sign in this interval.
For $x \in [-1, 1]$,$x^2 - x - 6$ is negative (e.g.,at $x = 0$,$y = -6$).
Thus,the area is $A = \int_{-1}^{1} |x^2 - x - 6| \, dx = \int_{-1}^{1} -(x^2 - x - 6) \, dx$.
$A = - \left[ \frac{x^3}{3} - \frac{x^2}{2} - 6x \right]_{-1}^{1}$.
$A = - \left[ (\frac{1}{3} - \frac{1}{2} - 6) - (-\frac{1}{3} - \frac{1}{2} + 6) \right]$.
$A = - \left[ \frac{1}{3} - \frac{1}{2} - 6 + \frac{1}{3} + \frac{1}{2} - 6 \right]$.
$A = - \left[ \frac{2}{3} - 12 \right] = - \left[ \frac{2 - 36}{3} \right] = - \left[ -\frac{34}{3} \right] = \frac{34}{3}$ sq. units.

(D) The area $A$ is given by the integral of the absolute value of the function over the interval $[1, 3]$.
$A = \int_{1}^{3} |\sin(\pi x)| \, dx$.
Since $\sin(\pi x)$ changes sign at $x = 2$,we split the integral:
$A = \int_{1}^{2} |\sin(\pi x)| \, dx + \int_{2}^{3} |\sin(\pi x)| \, dx$.
In the interval $[1, 2]$,$\sin(\pi x) \leq 0$,so $|\sin(\pi x)| = -\sin(\pi x)$.
In the interval $[2, 3]$,$\sin(\pi x) \geq 0$,so $|\sin(\pi x)| = \sin(\pi x)$.
$A = \int_{1}^{2} -\sin(\pi x) \, dx + \int_{2}^{3} \sin(\pi x) \, dx$.
Evaluating the integrals:
$A = \left[ \frac{\cos(\pi x)}{\pi} \right]_{1}^{2} + \left[ -\frac{\cos(\pi x)}{\pi} \right]_{2}^{3}$.
$A = \frac{1}{\pi} (\cos(2\pi) - \cos(\pi)) - \frac{1}{\pi} (\cos(3\pi) - \cos(2\pi))$.
$A = \frac{1}{\pi} (1 - (-1)) - \frac{1}{\pi} (-1 - 1) = \frac{2}{\pi} + \frac{2}{\pi} = \frac{4}{\pi}$.
Thus,the correct option is $D$.

(B) The area $A$ is given by the integral of the absolute value of the function over the interval $[\frac{\pi}{2}, \frac{3\pi}{2}]$.
$A = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} |\cos x| \, dx$
In the interval $[\frac{\pi}{2}, \frac{3\pi}{2}]$,$\cos x$ is negative in the interval $(\frac{\pi}{2}, \frac{3\pi}{2}]$.
Thus,$|\cos x| = -\cos x$.
$A = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} -\cos x \, dx$
$A = -[\sin x]_{\frac{\pi}{2}}^{\frac{3\pi}{2}}$
$A = -[\sin(\frac{3\pi}{2}) - \sin(\frac{\pi}{2})]$
$A = -[-1 - 1]$
$A = -[-2] = 2$ sq. units.
Therefore,the correct option is $B$.

(A) The area $A$ of the region bounded by the curve $y = f(x)$,the $x$-axis,and the lines $x = a$ and $x = b$ is given by the formula $A = \int_{a}^{b} |f(x)| \, dx$.
Given the curve $y = -2\sqrt{x}$,the lines $x = 0$ and $x = 1$,and the $x$-axis $(y = 0)$,the area is:
$A = \int_{0}^{1} |-2\sqrt{x}| \, dx$
$A = \int_{0}^{1} 2\sqrt{x} \, dx$
$A = 2 \int_{0}^{1} x^{1/2} \, dx$
$A = 2 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1}$
$A = 2 \times \frac{2}{3} [x^{3/2}]_{0}^{1}$
$A = \frac{4}{3} (1^{3/2} - 0^{3/2})$
$A = \frac{4}{3}$ sq. units.
Thus,the correct option is $A$.

(C) The area $A$ of the region bounded by the curve $y = f(x)$,the $X$-axis,and the lines $x = a$ and $x = b$ is given by the integral $A = \int_{a}^{b} y \, dx$.
Given the curve $y = x^2 + 2$,the limits $x = 1$ and $x = 2$,the area is:
$A = \int_{1}^{2} (x^2 + 2) \, dx$
Integrating the function:
$A = [\frac{x^3}{3} + 2x]_{1}^{2}$
Substituting the upper and lower limits:
$A = (\frac{2^3}{3} + 2(2)) - (\frac{1^3}{3} + 2(1))$
$A = (\frac{8}{3} + 4) - (\frac{1}{3} + 2)$
$A = (\frac{8 + 12}{3}) - (\frac{1 + 6}{3})$
$A = \frac{20}{3} - \frac{7}{3} = \frac{13}{3}$
Thus,the area is $\frac{13}{3}$ sq. units.

(B) The given equation of the ellipse is $25x^2 + 16y^2 = 400$.
Dividing both sides by $400$,we get:
$\frac{25x^2}{400} + \frac{16y^2}{400} = 1$
$\frac{x^2}{16} + \frac{y^2}{25} = 1$
Comparing this with the standard equation $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$,we have $a^2 = 25$ and $b^2 = 16$.
Thus,$a = 5$ and $b = 4$.
The area of an ellipse is given by the formula $A = \pi ab$.
Substituting the values,$A = \pi \times 5 \times 4 = 20\pi$ sq. units.

(D) The area $A$ is given by the integral of the absolute value of the function:
$A = \int_{0}^{3\pi} |\cos x| \, dx$
Since the function $\cos x$ changes sign at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ and $x = \frac{5\pi}{2}$,we split the integral:
$A = \int_{0}^{\pi/2} \cos x \, dx - \int_{\pi/2}^{3\pi/2} \cos x \, dx + \int_{3\pi/2}^{5\pi/2} \cos x \, dx - \int_{5\pi/2}^{3\pi} \cos x \, dx$
Evaluating each part:
$|\sin x|_{0}^{\pi/2} = |1 - 0| = 1$
$|\sin x|_{\pi/2}^{3\pi/2} = |-1 - 1| = |-2| = 2$
$|\sin x|_{3\pi/2}^{5\pi/2} = |1 - (-1)| = |2| = 2$
$|\sin x|_{5\pi/2}^{3\pi} = |0 - 1| = |-1| = 1$
Summing these values: $A = 1 + 2 + 2 + 1 = 6$ sq. units.
Thus,the correct option is $D$.

(A) The given curve is $5y = 5 - x$,which can be written as $y = 1 - \frac{x}{5}$.
To find the area bounded by the curve,the $X$-axis,and the lines $x = 1$ and $x = 4$,we use the definite integral:
$\text{Area} = \int_{1}^{4} y \, dx$
$\text{Area} = \int_{1}^{4} (1 - \frac{x}{5}) \, dx$
Integrating with respect to $x$:
$\text{Area} = [x - \frac{x^2}{10}]_{1}^{4}$
Substitute the limits:
$\text{Area} = (4 - \frac{16}{10}) - (1 - \frac{1}{10})$
$\text{Area} = (4 - 1.6) - (1 - 0.1)$
$\text{Area} = 2.4 - 0.9 = 1.5$
Thus,the area is $1.5$ sq. units,which is $\frac{3}{2}$ sq. units.

(D) The given curve is $y = |x - 3|$.
For the interval $x \in [0, 2]$,the expression $(x - 3)$ is always negative,so $|x - 3| = -(x - 3) = 3 - x$.
The area $A$ is given by the integral $\int_{0}^{2} (3 - x) \, dx$.
Evaluating the integral: $A = [3x - \frac{x^2}{2}]_{0}^{2}$.
Substituting the limits: $A = (3(2) - \frac{2^2}{2}) - (3(0) - \frac{0^2}{2})$.
$A = (6 - 2) - 0 = 4$ sq. units.
Therefore,the correct option is $D$.

(B) The equation of the circle is $x^2 + y^2 = 16$, which can be written as $x^2 + y^2 = 4^2$. This is a circle centered at the origin $(0, 0)$ with a radius $r = 4$.
In the first quadrant, the equation of the circle is $y = \sqrt{16 - x^2}$.
The region is bounded by $x = 0$ and $x = 4$ in the first quadrant.
The area $A$ is given by the integral: $A = \int_{0}^{4} y \, dx = \int_{0}^{4} \sqrt{16 - x^2} \, dx$.
Using the standard integral formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a}) + C$, we get:
$A = [\frac{x}{2} \sqrt{16 - x^2} + \frac{16}{2} \sin^{-1}(\frac{x}{4})]_{0}^{4}$.
$A = [\frac{4}{2} \sqrt{16 - 16} + 8 \sin^{-1}(1)] - [0 + 8 \sin^{-1}(0)]$.
$A = [0 + 8(\frac{\pi}{2})] - [0 + 0] = 4\pi$.
Thus, the area is $4\pi$ sq. units.

(B) To find the area bounded by the curve $y = x^2 - x$ and the $X$-axis,we first find the intersection points with the $X$-axis by setting $y = 0$.
$x^2 - x = 0 \implies x(x - 1) = 0$.
So,the intersection points are $x = 0$ and $x = 1$.
The curve $y = x^2 - x$ lies below the $X$-axis in the interval $(0, 1)$ because for any $x \in (0, 1)$,$x^2 < x$,so $y < 0$.
The area $A$ is given by the integral:
$A = \left| \int_{0}^{1} (x^2 - x) \, dx \right|$
$A = \left| \left[ \frac{x^3}{3} - \frac{x^2}{2} \right]_{0}^{1} \right|$
$A = \left| \left( \frac{1}{3} - \frac{1}{2} \right) - (0 - 0) \right|$
$A = \left| -\frac{1}{6} \right| = \frac{1}{6}$ sq. units.

(B) Given the curve equation $2y = -x + 8$,we can express $y$ as $y = \frac{-x + 8}{2} = -\frac{1}{2}x + 4$.
To find the area bounded by the curve,the $X$-axis,and the lines $x = 3$ and $x = 5$,we use the definite integral:
$\text{Area} = \int_{3}^{5} y \, dx = \int_{3}^{5} (-\frac{1}{2}x + 4) \, dx$.
Integrating the function:
$\int (-\frac{1}{2}x + 4) \, dx = [-\frac{1}{2} \cdot \frac{x^2}{2} + 4x] = [-\frac{x^2}{4} + 4x]$.
Now,applying the limits from $3$ to $5$:
$\text{Area} = [-\frac{5^2}{4} + 4(5)] - [-\frac{3^2}{4} + 4(3)]$.
$\text{Area} = [-\frac{25}{4} + 20] - [-\frac{9}{4} + 12]$.
$\text{Area} = [-\frac{25}{4} + \frac{80}{4}] - [-\frac{9}{4} + \frac{48}{4}]$.
$\text{Area} = \frac{55}{4} - \frac{39}{4} = \frac{16}{4} = 4$.
Thus,the area is $4$ sq. units.
Therefore,the correct option is $B$.

(A) The area $A$ bounded by the curve $y = f(x)$,the $X$-axis,and the lines $x = a$ and $x = b$ is given by the integral $A = \int_{a}^{b} |f(x)| \, dx$.
Given the curve $y = 2x^2$,the $X$-axis,and the line $x = 1$,the region is bounded from $x = 0$ to $x = 1$.
Thus,the area $A = \int_{0}^{1} 2x^2 \, dx$.
Evaluating the integral: $A = 2 \left[ \frac{x^3}{3} \right]_{0}^{1}$.
$A = 2 \left( \frac{1^3}{3} - \frac{0^3}{3} \right) = 2 \left( \frac{1}{3} \right) = \frac{2}{3}$ sq. units.
Therefore,the correct option is $A$.

(D) The area $A$ of the region bounded by the curve $y = f(x)$,the $X$-axis,and the lines $x = a$ and $x = b$ is given by the integral $A = \int_{a}^{b} |f(x)| \, dx$.
Here,$f(x) = 5 \sin x$,$a = 0$,and $b = \frac{\pi}{2}$.
Since $\sin x \geq 0$ for $0 \leq x \leq \frac{\pi}{2}$,we have:
$A = \int_{0}^{\frac{\pi}{2}} 5 \sin x \, dx$
$A = 5 [-\cos x]_{0}^{\frac{\pi}{2}}$
$A = 5 [-\cos(\frac{\pi}{2}) - (-\cos(0))]$
$A = 5 [0 + 1]$
$A = 5 \times 1 = 5$ sq. units.
Thus,the correct option is $D$.

(B) To find the area of the region bounded by the curve $y=x$ and the lines $x=1$ and $x=10$,we use the definite integral:
Area $= \int_{a}^{b} y \, dx$
Here,$a = 1$,$b = 10$,and $y = x$.
Area $= \int_{1}^{10} x \, dx$
$= \left[ \frac{x^2}{2} \right]_{1}^{10}$
$= \frac{10^2}{2} - \frac{1^2}{2}$
$= \frac{100}{2} - \frac{1}{2}$
$= \frac{99}{2} \text{ sq. units.}$

(C) To find the area bounded by the curve $y = x^2 - x - 6$ and the $X$-axis,we first find the points where the curve intersects the $X$-axis by setting $y = 0$.
$x^2 - x - 6 = 0$
$(x - 3)(x + 2) = 0$
So,the intersection points are $x = -2$ and $x = 3$.
The area $A$ is given by the integral of the absolute value of the function from $x = -2$ to $x = 3$:
$A = \int_{-2}^{3} |x^2 - x - 6| \, dx$
Since the parabola opens upward and lies below the $X$-axis between its roots,the value of $y$ is negative in this interval.
$A = - \int_{-2}^{3} (x^2 - x - 6) \, dx$
$A = - [\frac{x^3}{3} - \frac{x^2}{2} - 6x]_{-2}^{3}$
Evaluating the definite integral:
At $x = 3$: $(\frac{27}{3} - \frac{9}{2} - 18) = 9 - 4.5 - 18 = -13.5$
At $x = -2$: $(\frac{-8}{3} - \frac{4}{2} + 12) = -2.666 - 2 + 12 = 7.333$
$A = - (-13.5 - 7.333) = - (-20.833) = 20.833$
Converting to fraction: $20.833 = \frac{125}{6}$ sq. units.

(D) The area $A$ of the region bounded by the curve $y = \cos x$,the $X$-axis,and the lines $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$ is given by the integral:
$A = \int_{-\pi/2}^{\pi/2} |\cos x| \, dx$
Since $\cos x \geq 0$ for $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,we have:
$A = \int_{-\pi/2}^{\pi/2} \cos x \, dx$
$A = [\sin x]_{-\pi/2}^{\pi/2}$
$A = \sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2})$
$A = 1 - (-1) = 1 + 1 = 2$
Thus,the area is $2$ sq. units.

(C) The area $A$ is given by the definite integral of the function $y = |x-5|$ from $x=5$ to $x=6$.
Since $x \ge 5$ in the interval $[5, 6]$,we have $|x-5| = x-5$.
Therefore,$A = \int_{5}^{6} (x-5) \, dx$.
Let $u = x-5$,then $du = dx$. When $x=5, u=0$ and when $x=6, u=1$.
$A = \int_{0}^{1} u \, du = \left[ \frac{u^2}{2} \right]_{0}^{1} = \frac{1}{2} - 0 = 0.5$ sq. units.
Thus,the correct option is $C$.

(D) The region is bounded by $y=x^2$ and $y=4$.
First,find the points of intersection by setting $x^2 = 4$,which gives $x = \pm 2$.
The area $A$ is given by the integral of the upper curve minus the lower curve from $x = -2$ to $x = 2$:
$A = \int_{-2}^{2} (4 - x^2) dx$
Since the function is symmetric about the $y$-axis,$A = 2 \int_{0}^{2} (4 - x^2) dx$.
$A = 2 [4x - \frac{x^3}{3}]_{0}^{2}$
$A = 2 [(4(2) - \frac{2^3}{3}) - (0)]$
$A = 2 [8 - \frac{8}{3}] = 2 [\frac{24-8}{3}] = 2 [\frac{16}{3}] = \frac{32}{3}$ sq. units.

(C) Given the equation of the curve $x+2y=8$,we can express $y$ in terms of $x$ as:
$2y = 8 - x$
$y = \frac{8-x}{2} = 4 - \frac{x}{2}$
To find the area bounded by the curve,the $X$-axis,and the lines $x=1$ and $x=5$,we use the definite integral:
$\text{Area} = \int_{1}^{5} y \, dx$
$\text{Area} = \int_{1}^{5} (4 - \frac{x}{2}) \, dx$
Integrating term by term:
$\text{Area} = [4x - \frac{x^2}{4}]_{1}^{5}$
Evaluating at the limits:
$\text{Area} = (4(5) - \frac{5^2}{4}) - (4(1) - \frac{1^2}{4})$
$\text{Area} = (20 - \frac{25}{4}) - (4 - \frac{1}{4})$
$\text{Area} = 20 - 6.25 - 4 + 0.25$
$\text{Area} = 16 - 6 = 10$
Thus,the area is $10$ sq. units.
The correct option is $C$.

(D) The given curve is $y = |x - 5|$.
Since the interval is $x \in [0, 2]$,we have $x - 5 < 0$,so $|x - 5| = -(x - 5) = 5 - x$.
The area $A$ is given by the integral:
$A = \int_{0}^{2} (5 - x) \, dx$
$A = [5x - \frac{x^2}{2}]_{0}^{2}$
$A = (5(2) - \frac{2^2}{2}) - (0 - 0)$
$A = 10 - 2 = 8$ sq. units.
Thus,the correct option is $D$.

(D) The given equation is $y = 2 \sqrt{1 - x^2}$.
Squaring both sides,we get $y^2 = 4(1 - x^2)$,which implies $x^2 + \frac{y^2}{4} = 1$.
This is the equation of an ellipse with semi-major axis $a = 2$ (along the $Y$-axis) and semi-minor axis $b = 1$ (along the $X$-axis).
Since the curve is $y = 2 \sqrt{1 - x^2}$,it represents the upper half of the ellipse.
The area bounded by the curve and the $X$-axis is the area of the upper semi-ellipse.
The area of the full ellipse is given by $A = \pi ab = \pi \times 1 \times 2 = 2 \pi$.
Therefore,the area of the upper semi-ellipse is $\frac{1}{2} \times 2 \pi = \pi$ sq. units.

(C) To find the area bounded by the curve $y = 2x - x^2$ and the $X$-axis,we first find the points of intersection with the $X$-axis by setting $y = 0$.
$2x - x^2 = 0$
$x(2 - x) = 0$
So,$x = 0$ and $x = 2$.
The area $A$ is given by the integral of the curve from $x = 0$ to $x = 2$:
$A = \int_{0}^{2} (2x - x^2) \, dx$
$A = [x^2 - \frac{x^3}{3}]_{0}^{2}$
$A = (2^2 - \frac{2^3}{3}) - (0^2 - \frac{0^3}{3})$
$A = 4 - \frac{8}{3}$
$A = \frac{12 - 8}{3} = \frac{4}{3}$ sq. units.

(C) The area $A$ is given by the integral of the function $y = \cot x$ from $x = \frac{\pi}{4}$ to $x = \frac{\pi}{2}$.
$A = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot x \, dx$
We know that $\int \cot x \, dx = \log |\sin x| + C$.
Applying the limits:
$A = [\log |\sin x|]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$
$A = \log |\sin(\frac{\pi}{2})| - \log |\sin(\frac{\pi}{4})|$
$A = \log(1) - \log(\frac{1}{\sqrt{2}})$
Since $\log(1) = 0$ and $\log(\frac{1}{\sqrt{2}}) = \log(2^{-1/2}) = -\frac{1}{2} \log 2$:
$A = 0 - (-\frac{1}{2} \log 2) = \frac{1}{2} \log 2$.
Thus,the correct option is $C$.

(C) The area $A$ of the region bounded by the curve $y = f(x)$,the $X$-axis,and the lines $x = a$ and $x = b$ is given by the integral:
$A = \int_{a}^{b} |y| \, dx$
Here,$y = \tan x$,$a = 0$,and $b = \frac{\pi}{4}$.
Since $\tan x \geq 0$ in the interval $[0, \frac{\pi}{4}]$,we have:
$A = \int_{0}^{\frac{\pi}{4}} \tan x \, dx$
The integral of $\tan x$ is $\ln|\sec x|$:
$A = [\ln|\sec x|]_{0}^{\frac{\pi}{4}}$
$A = \ln|\sec(\frac{\pi}{4})| - \ln|\sec(0)|$
$A = \ln(\sqrt{2}) - \ln(1)$
$A = \ln(2^{1/2}) - 0$
$A = \frac{1}{2} \ln 2$
Thus,the correct option is $C$.